Strictly speaking, yes. O-notation describes the growth of a function as the argument tends to infinity. Formally, the statement "f(x) is O(g(x))" means that there's some point x_0 such that f(x) is less than a constant factor times g(x) for any x > x_0.

This is clearly the case for f(n) = 1, g(n) = lg(n), hence the constant function is O(lg(n)).

An algorithm can be O of different g(x)s, depending on the properties of the input. For example, the runtime of a naive quicksort implementation (which always chooses the leftmost value as a pivot) is O(n^2) if the input list is sorted, while it has runtime O(n*lg(n)) on average and in the best case (where the pivot is always the median of the section of the list being partitioned).